Fix $n > 1$ and let $[n] := \{ 1, 2, \dots, n \}$. Which probability mass function (PMF) over $[n]$ has the largest $2$-norm?
Doodling for the cases $n \in \{2,3\}$ does suggest that
the maximal $2$-norm is attained when the support of the PMF is a singleton, i.e., at the "corners" of the probability simplex
the minimal $2$-norm is attained when the PMF is uniform over $[n]$
In essence, we have the following (non-convex) quadratic program (QP)
$$ \begin{array}{ll} \underset {{\bf x} \in \Bbb R^n} {\text{maximize}} & \| {\bf x} \|_2^2 \\ \text{subject to} & {\bf 1}_n^\top {\bf x} = 1 \\ & {\bf x} \geq {\bf 0}_n \end{array} $$
One way of getting rid of ${\bf x} \geq {\bf 0}_n$ is to introduce $n$ new variables $y_i^2 := x_i$ and rewrite the QP above as follows
$$ \begin{array}{ll} \underset {{\bf y} \in \Bbb R^n} {\text{maximize}} & \sum\limits_{i=1}^n y_i^4\\ \text{subject to} & \sum\limits_{i=1}^n y_i^2 = 1 \end{array} $$
where there is a single equality constraint. We define the Lagrangian
$$ \mathcal L ({\bf y}, \mu) := \frac14 \sum\limits_{i=1}^n y_i^4 - \frac{\mu}{2} \left( \sum\limits_{i=1}^n y_i^2 - 1 \right) $$
Differentiating and finding where the partial derivatives vanish, we obtain
$$ \begin{aligned} y_1 \left( y_1^2 - \mu \right) &= 0 \\ &\vdots \\ y_n \left( y_n^2 - \mu \right) &= 0 \\ \sum\limits_{i=1}^n y_i^2 &= 1 \end{aligned} $$
Note that $y_i = 0$ or $y_i^2 = \mu$. Let $\operatorname{card} ({\bf y})$ denote the cardinality of the support, i.e., the number of non-zero entries of $\bf y$. Hence,
$$\sum\limits_{i=1}^n y_i^2 = \mu \operatorname{card} ({\bf y}) = 1$$
and, thus, $\color{blue}{x_i \in \left\{ 0, \dfrac{1}{\operatorname{card} ({\bf x})} \right\}}$. Note that $\| {\bf x} \|_2^2 = \dfrac{1}{\operatorname{card} ({\bf x})}$, which is maximal when $\color{blue}{\operatorname{card} ({\bf x}) = 1}$.
Is this correct? If so, is there a more elegant way of showing that ${\bf x}_{\max} \in \{ {\bf e}_1, {\bf e}_2, \dots, {\bf e}_n \}$?
Related
